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Question
Evaluate: `int "dx"/("9x"^2 - 25)`
Solution
Let I = `int "dx"/("9x"^2 - 25)`
`= int 1/(9 ("x"^2 - 25/9))` dx
`= 1/9 int 1/("x"^2 - (5/3)^2)` dx
`= 1/9 * 1/(2 * 5/3) log |("x" - 5/3)/("x" + 5/3)|` + c
∴ I = `1/30 log |(3"x" - 5)/("3x" + 5)|` + c
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